Problem: Prove that $(E(X) +E(\frac{1}{X})) \geq 2$ where, $X$ is a non-negative random variable.
My approach:
For any non-negative random variable $Y$ we have
$$ E((Y-\frac{1}{Y})^2)\geq 0 \implies E(Y^2)+E(\frac{1}{Y^2}) \geq 2 \tag 1 $$
Let $Y =\sqrt{X}$, which will also be a non-negative random variable.
Substituting $Y$ in $(1)$, we get
$$
E(X)+E(\frac{1}{X}) \geq 2
$$
Is there any fault in my proof?
Your proof works quite all right!
Arthur already gave the proper comment, this is just an alternative way to prove the statement - maybe a bit more straight forward, to elaborate a bit:
We want to show for $X$ positive, that $$ (E(X) +E(\frac{1}{X})) \geq 2 $$ holds, but this is due linearity of the expectation operator equivalent to $$ E(X +\frac{1}{X}) \geq 2 $$ now we use the fact, that for $x>0$ it holds that $$ x+\frac1x\geq2 \tag 1 $$ because $$ x+\frac1x\geq2\Leftrightarrow x^2-2x+1=(x-1)^2\geq0 $$ which is of course a true statement. Now we just use $(1)$ and see $$ (E(X) +E(\frac{1}{X}))=E(X +\frac{1}{X})\geq E(2)=2 $$ and we are done.